I've been reading a paper by R.Ramakrishna about lifting Galois representations and at some point he uses the fact that a representation with big image has trivial first cohomology group.

In other words, what i've been trying to prove (and failing at) is that given $p>5$ a prime, if $M$ is the group of trace zero $2\times 2$ matrices over $\mathbb{F}_p$ then:

$H^1(SL_2(\mathbb{F}_p),M) = 0$

where the action is given by conjugation.

My first approach was calculating the group $H^1(S_p,M)$, where $S_p$ is a $p$-sylow subgroup of $SL_2(\mathbb{F}_p)$. If it were trivial then the one that I want should be, because $M$ is a $p$-group, but it turns out it's not! (I think it's one dimensional).

Maybe there is some cohomological-theoretic reason I'm missing, thanks.

As Tom suggests, in this kind of Lie-type finite group the most useful cohomology comparison is with a Borel subgroup (normalizer of the Sylow) rather than with its unipotent radical (the Sylow $p$-group).
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Jim HumphreysOct 13 '11 at 13:51

Also, adding a tag gr.group-theory would be helpful. Some group theorists might suggest approaches to this problem which don't use too much extra machinery. I tend to find the algebraic group approach most natural, of course, but not everyone would.
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Jim HumphreysOct 14 '11 at 20:42

I think the suggestion works out well, I have to check my calculations but it seems that the restriction to the Borel is trivial. Also added the group theory tag. Thanks.
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mcampoOct 14 '11 at 21:47

3 Answers
3

As Olivier's answer indicates, a direct approach to the computation using classical techniques of Sah and others is possible in a small isolated case like this. But it's a good idea to be aware of several different approaches to computations of finite group cohomology (or more generally, Ext functors). Most such computations are extremely hard to carry out explicitly, but sometimes it's enough to work in the setting of finite group cohomology with no added machinery. There are also some powerful techniques coming from algebraic topology (book of Adem-Milgram, etc.). For the groups $SL(2,p^n)$, a detailed study of the cohomology ring by Jon Carlson in Proc. LMS 47 (1983) shows how difficult the whole subject can be.

In the question here (and often in the study of Galois representations) you are dealing with a family of finite groups of Lie type, where it's often helpful to think about comparisons with the cohomology of the ambient algebraic group. Here there is an extensive literature, summarized concisely but with lots of references in my 2005 book Modular Representations of Finite Groups of Lie Type (LMS Lecture Note Series 326): work of Cline-Parshall-Scott and their collaborators, etc. In general the question asked has a transparent answer if the "top" or "bottom" layers of the projective cover of the trivial module can be worked out explicitly. This isn't at all easy in general, but for $SL(2,p^n)$ there is a fairly direct comparison with related modules for the algebraic group: see the paper by Henning Andersen et al. in Proc. LMS 46 (1983).

P.S. One advantage of looking directly at the projective cover of the trivial module (as in the last-cited paper) is that it explains why $p=5$ causes trouble: the two highest weights involved are $\lambda=0$ (trivial module) and $\mu =2$ (adjoint module), while the condition for existence of a nonsplit extension between them is easily seen from the algebraic group comparison to be $\lambda+\mu = p-3$.

(Sah's lemma) Let $G$ be a group, $M$ a $G$-representation and $g\in Z(G)$. Then $x\mapsto (g-1)x$ is the zero map on $H^{1}(G,M)$.

The proof is a computation of $(g-1)f(h)$ using the cocycle relation and the fact that $f(h)=f(ghg^{-1})$. Applying the lemma to $\operatorname{SL}_{2}(\mathbb F_{p})$, $M$ and a non-trivial homotethy, for instance $-Id$, gives the result.

But in this case all the elements which lie in the center act trivially on M, since they are diagonal matrices and the action is by conjugation. I think the lemma doesn't help here.
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mcampoOct 13 '11 at 21:18

You're right: I completely overlooked that, though it was stated clearly in your post.
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OlivierOct 14 '11 at 5:06

Aside remark: there are statement(s) similar to Sah's lemma for cohomology of Lie algebras, and, probably, it can be proved from general principles and in higher cohomology degrees (see papers of Rolf Farnsteiner around 1987-1990).
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Pasha ZusmanovichOct 21 '11 at 17:30

This question is Lemma 2.48 in "Fermat's Last Theorem" by Darmon, Diamond, and Taylor. Here is there argument (not quite verbatim, but almost...)

Let $U\subset B \subset G = SL_2(\mathbb{F}_p)$ be the subgroups of unipotent and upper triangular matrices, respectively. Since $[G:B]=p+1$ is prime to $p$, the group $H^1(G/B, M) = 0$ so that the restriction map $H^1(G,M) \to H^1(B,M)$ is injective. Similarly, $[B:U] = p-1$ so that $H^1(B,M) \simeq H^1(U,M)^{B/U}$.

Now, there is a filtration $$0 \subset M'' \subset M' \subset M$$ where $M'$ are the upper triangular trace zero matrices and $M''$ are those in $M'$ with zeros on the diagonal. By comparing the long exact sequences of Galois cohomology which arise from the short exact sequences $$ 0 \to M'\to M\to M/M'\to 0$$ and $$0 \to M''\to M'\to M'/M''\to 0,$$ you can see that $H^1(U,M)$ injects into $H^1(U,M/M')$. This last group is just $\mathrm{Hom}(\mathbb{F}_p,\mathbb{F}_p)$ and you can just show directly then that $\mathrm{Hom}(\mathbb{F}_p,\mathbb{F}_p)^{B/U}=0$.

When you are comparing the long exact sequences, you can get most of the information you need by computing the dimension of $H^i(U,X)$ as an $\mathbb{F}_p$-vector space (here $X$ is $M,\ M',\ M''$, or one of the quotients). You can compute the dimensions by just writing down the action of U on M in terms of matrix multiplication.